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Shalin (and Guy),
thank you so much for the explanation and the article, it is exactly what
I was looking for. It now makes sense.
I have to disclose that there may be one additional factor that
contributes to image degradation with the 20x objective (and has
something to do with the large diameter of the BFP) - I looked at the
DIC prism and discovered that there is an oil droplet close to he edge of
the prism. Generous application of immersion oil by our users must have
caused the immersion oil to run down, seep through the thread and drip
on the prism. So I think the oil droplet would degrade image more
severely for objectives with large BFP.
Now off to do some cleaning.
Stan
Dr. Stanislav Vitha
Microscopy and Imaging Center
Texas A&M University
BSBW 119
College Station, TX 77843-2257
http://microscopy.tamu.edu
On Sun, 10 Apr 2011 22:31:06 +0800, Shalin Mehta
<[log in to unmask]> wrote:
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>*****
>
>Hello Stan,
>
>On Fri, Apr 8, 2011 at 3:28 PM, Stanislav Vitha <[log in to unmask]>
wrote:
>>
>> My main question was not why I get degradation of resolution in the
>> fluorescence channel (that I expected), but rather why is the
degradation
>> worse for a lower-resolution objective (20x/0.85 oil imm versus
100x/1.4 oil
>> imm). The "image doubling" due to the shear of the objective DIC
prism should
>> be more noticeable with the high-resolution lens
>
>I happened to have gone through this exercise when measuring the
shear
>in our DIC setup.
>We tried to recap what we understood in the appendix of our paper
>(http://www.mshalin.com/blog/wp-
content/uploads/2010/09/mehtaao2010-samplefreecalibrationdic.pdf).
>
>Here is an attempt at understanding the specific case that you
described:
>
>Nomarski prism is an angular polarizing beam-splitting devices. i.e.,
>it splits the incoming light into two polarizations that travel at
>different angles. The angular split caused by the Nomarski prism
>(placed in the back focal plane) amounts to spatial shear in the
>specimen plane. The relationship between angular shear (a) and spatial
>shear (s) is rather simple: s=tan(a)*f. Where f is the focal length of
>the objective. This fact can be seen easily from the geometry of fig.
>11(d) in above paper. Now the image doubling is caused when the
shear
>is large with respect to the resolution. To judge the doubling, we
>should normalize both sides of the above equation by lambda/NA.
Then,
>we have s/(lambda/NA) = tan(a) * f/(lambda/NA). Therefore, when
>s/(lambda/NA) or s*NA is large (for given wavelength) we will have
>image doubling.
>
>In the Olympus setup there is only one DIC prism on objective side,
>i.e., the angular shear is the same for all objectives. Therefore, the
>objectives for which f*NA is large will give rise to larger s*NA and
>hence image doubling.
>
>The product of focal length and NA is large for 20X 0.85 than for
>100X1.4. I compute the product of focal length and NA below.
>
>Objective's magnification (M) = tube length (ft)/ objective's focal
length (f).
>Therefore, f=ft/M.
>Therefore, f*NA = ft*(NA/M).
>For, 20X 0.85: f*NA= 0.0425*ft.
>For, 100x 1.4: f*NA= 0.014*ft.
>
>So the 'image doubling' for 20X 0.85 is almost 3 times more than for
100x 1.4.
>
>Now, it is interesting to see how this relates with Guy's intuitive
observation.
>
>NA of the objective = radius of the objective BFP (d)/ objective focal
>length (f).
>So, r=f*NA. As we have seen above, the size of the BFP (ie., f*NA)
>for 20X 0.85 is 3 times larger than for 100x1.4. So as Guy says, rays
>collected by 20X 0.85 pass through larger (3 times larger) surface of
>the Wollaston prism.
>
>Best
>Shalin
>
>http://mshalin.com
>
>Research Associate
>Bioimaging Lab, Block-E3A, #7-10
>Div of Bioengineering, NUS Singapore 117574
>(M) +65-84330724
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